this post was submitted on 01 Dec 2023
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cross-posted from: https://programming.dev/post/6660679

It's about asking, "how does this algorithm behave when the number of elements is significantly large compared to when the number of elements is orders of magnitude larger?"

Big O notation is useless for smaller sets of data. Sometimes it's worse than useless, it's misguiding. This is because Big O is only an estimate of asymptotic behavior. An algorithm that is O(n^2) can be faster than one that's O(n log n) for smaller sets of data (which contradicts the table below) if the O(n log n) algorithm has significant computational overhead and doesn't start behaving as estimated by its Big O classification until after that overhead is consumed.

#computerscience

Image Alt Text:

"A graph of Big O notation time complexity functions with Number of Elements on the x-axis and Operations(Time) on the y-axis.

Lines on the graph represent Big O functions which are are overplayed onto color coded regions where colors represent quality from Excellent to Horrible

Functions on the graph:
O(1): constant - Excellent/Best - Green
O(log n): logarithmic - Good/Excellent - Green
O(n): linear time - Fair - Yellow
O(n * log n): log linear - Bad - Orange
O(n^2): quadratic - Horrible - Red
O(n^3): cubic - Horrible (Not shown)
O(2^n): exponential - Horrible - Red
O(n!): factorial - Horrible/Worst - Red"

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[–] scrubbles@poptalk.scrubbles.tech 17 points 11 months ago (1 children)

This is why I only use bogo sort

[–] ericjmorey@programming.dev 8 points 11 months ago (2 children)

With a small enough data set, bogo sort will perform just as well as an O(1) algorithm for sorting for both ascending and descending order!

[–] scrubbles@poptalk.scrubbles.tech 3 points 11 months ago (1 children)
[–] ericjmorey@programming.dev 4 points 11 months ago

Use it or lose it, right?

[–] mryessir@lemmy.sdf.org 3 points 11 months ago
[–] off_brand_@beehaw.org 16 points 11 months ago (2 children)

Yup, it's why O(N+10) and even O(2N) are effectively the same as O(N) on your CS homework. Speaking too generally, once you're dithering over the efficiency of an algorithm processing a 100-item dataset you've probably gone too far in weeds. And optimizations can often lead to messy code for not a lot of return.

That's mostly angled at new grads (or maybe just at me when I first started). You've probably got bigger problems to solve than shaving a few ms from the total runtime of your process.

[–] reboot6675@sopuli.xyz 16 points 11 months ago

Just a little nitpicking, O(N+10) and O(2N) are not "effectively" the same as O(N). They are truly the same.

[–] HarkMahlberg@kbin.social 7 points 11 months ago (3 children)

Aren't the most commonly accepted sorting algorithms O(nlog(n))? Quicksort? Mergesort? Those are considered bad?

[–] cawifre@beehaw.org 11 points 11 months ago

I mean, it is entirely reasonable that "bad" is the best performance you can hope for while sorting an entire set of generally comparable items.

If you can abuse special knowledge about the data being sorted then you can get better performance with things like radix sort, but in general it just takes a lot of work to compare them all even if you are clever to avoid wasted effort.

[–] Kajo@beehaw.org 8 points 11 months ago

Yeah, you're right, it doesn't make sense to say that O(f(n)) is good or bad for any algorithm. It must be compared to the complexity of other algorithms which solve the same problem in the same conditions.

[–] ericjmorey@programming.dev 1 points 11 months ago

The labels are from the perspective of viewing the space of all possible functions of element set size to operations, so they don't apply to any particular problem an algorithm is attempting to solve (that space is often smaller).

[–] chunkystyles@sopuli.xyz 4 points 11 months ago

I've been wrong about the performance of algorithms on tiny data sets before. It's always best to test your assumptions.

[–] wifienyabledcat@beehaw.org 3 points 11 months ago

god i wish i had this image when i was learning algorithms. i always just guessed for some of the exam questions 🙃